JEE Main 2026 — Center of Mass Momentum and Collision Question with Solution

Let the midpoint $p$ of the normal bisector be the origin $(0,0)$.

The triangle is isosceles with two sides of $10\text{ m}$ and an included angle of $120^{\circ }$. The altitude from the $15\text{ kg}$ mass bisects the angle into two $60^{\circ }$ angles.

The length of the altitude is $10\cos (60^{\circ })=5\text{ m}$.

Since $p$ is the midpoint of this altitude, the $15\text{ kg}$ mass is at a distance of $2.5\text{ m}$ above $p$, and the base is $2.5\text{ m}$ below $p$.

The distance of the $2\text{ kg}$ and $3\text{ kg}$ masses from the altitude is $10\sin (60^{\circ })=5\sqrt{3}\text{ m}$.

The coordinates of the masses with respect to $p$ are:
$m_1=15\text{ kg}$ at $(0,2.5)$
$m_2=2\text{ kg}$ at $(-5\sqrt{3},-2.5)$
$m_3=3\text{ kg}$ at $(5\sqrt{3},-2.5)$

The x-coordinate of the center of mass is:
$x_{cm}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3}=\frac{15(0)+2(-5\sqrt{3})+3(5\sqrt{3})}{15+2+3}=\frac{5\sqrt{3}}{20}=\frac{\sqrt{3}}{4}$

The y-coordinate of the center of mass is:
$y_{cm}=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3}{m_1+m_2+m_3}=\frac{15(2.5)+2(-2.5)+3(-2.5)}{20}=\frac{37.5-12.5}{20}=\frac{25}{20}=1.25$

The position of the center of mass is $\left(\frac{\sqrt{3}}{4},1.25\right)$.

Answer: $\left(\frac{\sqrt{3}}{4},1.25\right)$